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Let $s_n=\sum_{k=1}^n\left<v,e_k\right>e_k$. Then $$\left<v,s_n\right>=\sum_{k=1}^n|\left<v,e_k\right>|^2$$ and $$\|s_n\|^2=\sum_{k=1}^n|\left<v,e_k\right>|^2.$$ Thus $$\|s_n\|^2=\left<v,s_n\right>$$ and hence $$\|s_n\|^2\le \|v\|\|s_n\|$$ So $\|s_n\|\le\|v\|$ and hence $\sum_{k=1}^n|\left<v,e_k\right>|^2\le\|v\|^2$. ...

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The following comes from Rudin's Functional Analysis 2ed. Theorem 4.25(a) Let $X$ be a Banach space, with $T\in\mathcal{B}(X)$ compact. If $\lambda\neq0$ then $\dim\mathcal{N}(T-\lambda I)=\dim X/\mathcal{R}(T-\lambda I),$ and both are finite. In fact, Rudin says a lot more, but this is all we need. By hypothesis, $I-T$ is onto, and therefore $\dim X/... 2 Regarding convergence and completeness: For$n\in N$let$f_n(x)=0$for$x\leq 1/2-1/(n+2)$and$f_n(x)=1$for$x\geq 1/2.$Let$f_n(x)$be linear for$x\in [1/2-1/(n+2),1/2].$Then$(f_n)_{n\in N}$is a Cauchy sequence with respect to the norm$\|f-g\|=[\int_0^1|f(x)-g(x)|^2\;dx]^{1/2}.$Let$h(x)=0$for$x\leq 1/2$and$h(x)=1$for$x>1/2.$The ... 0 Re : The proposer's comment to carmichael561's answer: Let$\{e_n:n\in N\}$be an orthonormal Hilbert-space basis for$H.$Let$h_1=e_1$and$h_2=e_1-e_2. $For$n\geq 3$let$h_n=e_1+2e_{n-1}-e_n.$Then$e_1$has two representations with respect to$\{h_n: n\in N\},$namely $$(i) \quad e_1=h_1.$$ $$(ii) \quad e_1=\sum_{n=1}^{\infty} h_n 2^{-n}.$$ 1 For a simple example to demonstrate the importance of the basis being orthonormal, consider$\mathbb{R}^2$with the standard inner product and the basis$h_1=(1,0)$and$h_2=(1,1)$. If$h=(0,1)$then $$h=-h_1+h_2$$ but$\langle h,h_1\rangle=0$. By the way, a Hilbert space basis of$H$is different from a basis of$H$as an abstract vector space (called a ... 3 To elaborate on the suggestion in the comments: Assume$A=\langle x, y \rangle \ne 0$. After replacing$y$with$\alpha\, y$, with non-zero$\alpha \in \mathbb C$, you can assume that$A$is (non-zero) real. At that point, square the inequality, and expand the lhs - you may take$c$to be real to find your desired contradiction. Hope this helps. 2 I think the intended question was something like: Given$p(x)\in\mathbb{C}[x]$with degree$n$, such that $$\int_{0}^{1}x^k p(x)\,dx=0$$ for every$k\in\{1,2,\ldots,n-1\}$, show that the roots of$p(x)$lie in the circle$|x|\leq 1$. We may notice that$p(x)$has$n+1$coefficients and we have$n-1$linear constraints on them, so the space of ... 3 The function$t\cdot \chi_{[-1,1]}(t)$is orthogonal to each$t^{2k}$in$L^2[-1,3],$hence is orthogonal to the the linear span of$\{t^{2k} : k=0,1,\dots \}$in$L^2[-1,3],$hence is orthogonal to the closure of this linear span in$L^2[-1,3].$Therfore this closure cannot be all of$L^2[-1,3].$2 Let$A$denote the linear span of$\{t^{2k}\}_{k \in \mathbb{N}}$. Then, if instead of$[-1,3]$the domain was$[1,3]$instead, we could use the Stone-Weierstrass theorem to conclude that$A$is dense ins$C([1,3])$and thus in$L^2([1,3])$since$t^2$separates the points of$[1,3]$and$1 \in A$. However, for$[-1,3]$, we have$t^{2k}(-1) = t^{2k}(1)$... 1 \begin{eqnarray} \langle T_z^*(\alpha),x \rangle_{H}&=&\langle \alpha , T_z(x)\rangle_\mathbb{K}\\ &=& \alpha \overline{T_z(x)}\\ &=& \alpha\langle z,x\rangle_H\\ &=& \langle \alpha z , x\rangle_H\\ \end{eqnarray} So $$T_z^*(\alpha)=\alpha z$$ It's clear now that$\|T_z\|=\|T_z^*\|=\|z\|$0 Let$x\in \mathbb R^n$be given. Set$u:=\sum_{i=1}^n x_iv_i \in H$. Then $$x^T(A_1-2\rho A_2+\rho A_3)x =\langle u, (I -2\rho B+\rho^2 B^2)u\rangle,$$ since$B$is self-adjoint and$\rho$is real. Now,$I -2\rho B+\rho^2 B^2 = (I-\rho B)^2$, and again by self-adjointness $$x^T(A_1-2\rho A_2+\rho A_3)x =\langle u, (I -2\rho B+\rho^2 B^2)u\rangle =\|(I-\... 2 So first of all, H^1 does not really have a restriction map, even to interior points, much less boundary points (except in one dimension, where there is a continuous embedding into C^0). The better way of thinking about this is to start out with smooth test functions, then look at the equation that you get and identify the solution space and the test ... 2 It is usual to assume some regularity property for \Gamma. For example, in Evan's PDE, if \Omega is bounded and has a C^1 boundary, then there is (bounded) trace operator$$T : H^1(\Omega) \to L^2(\Gamma)$$so that T u = u|_\Gamma if u \in H^1(\Omega) \cap C(\overline\Omega). It is also proved that$$\{ H^1(\Omega) : Tu = 0\} = H^1_0(\Omega),... 1 For the map$\Phi \colon H \to U'$,$x \mapsto \langle \cdot, x \rangle$we have$\ker \Phi = U^\perp$. Hence$\Phi$is injective if and only if$U^\perp = 0$. Because$H = U^\perp \oplus \overline{U}$we have$U^\perp = 0$if and only if$\overline{U} = H$, i.e. if$U$is dense. 1 In a Hilbert space, you have the Riesz Representation Theorem, which tells you that given any$f\in X'$, there exists$y\in X$such $$\tag{1}f(y)=\langle y,x\rangle,\ \ \ x\in X.$$ And this assignment is isometric so$X'$is isomorphic to$X$. In practice, one thinks that$X'=X$via the duality$(1)$. In summary, in the case of a Hilbert$X$space the ... 2 You have pointwise convergence of each term by the first condition. By the second condition condition there is a sequence converging to zero that majorizes the terms for arbitrary$ n $. This allows you to just split up the sum and show that both parts are close to zero. EDIT: It is only a little bit trickier than what you now have and than what I sketched ... 1 Note that the range of$K$is in a finite dimensional subspace $$\text{span} \{x, \cos x\},$$ thus$K$is a compact operator and has only point spectrum ($0$is a eigenvalue too). To look for eigenfunction, we need only restrict ourselve to the above subspace. Note that $$K (ax+b \cos x) = b\pi x + a\frac{2\pi^3}{3} \cos x,$$ This implies that $$... 1 Suppose Av=\lambda v. Then as you noted we have (Av)_j=\frac 1{2^{j-1}}(Av)_1=\frac \lambda{2^{j-1}}v_1 but also (Av)_j=\lambda v_j. Thus we have that \lambda v_j=\frac \lambda{2^{j-1}}v_1 so that v_j=\frac{v_1}{2^{j-1}}. For v_1, we have that \lambda v_1=(Av)_1=\sum_{n=1}^\infty\frac{v_1}{2^{n-1}2^n} which implies that \lambda=2\sum_{n=1}^\... 1 That is, taking j=i, \alpha_i (\lambda - \lambda_i) = 0, so you can't have \alpha_i \ne 0 unless \lambda_i = \lambda. And since some \alpha_i must be nonzero, you get your desired conclusion. 1 You have already shown that$$ \|A_nf\|_2\leq\frac {\|K\|_1\,\log n}{\sqrt n}\, \|f\|_2, $$which says that$$ \|A_n\|\leq\frac {\|K\|_1\,\log n}{\sqrt n}. $$0 There are at least two plausible interpretations for the equation$$ \frac{d}{dt} \langle\psi(t), A\psi(t)\rangle = \left\langle\frac{d\psi}{dt}, A\psi\right \rangle + \left\langle\psi, A\frac{d\psi}{dt}\right\rangle; \tag{1} $$not sure which, if either, the author intended. The time-dependent wave function \psi is taken to be L^{2}-valued, not ... 1 There may well be a simpler or more elegant solution. We may as well simplify the notation by assuming ||f_k||_\infty\le 1; since ||g_k||_2 is bounded let's assume ||g_k||_2\le 1 as well. Say \epsilon>0. Choose \delta>0 so m(E)<\delta implies$$\left(\int_E|\phi|^2\right)^{1/2}<\epsilon.$$Egoroff's Theorem shows that there exists ... 1 Let V\subseteq\mathbb{R}^n be the image of A, and let W\subseteq \mathbb{R}^n be an orthogonal complement thereof. Then we can write uniquely b=b_v+b_w, with b_v\in V and b_w\in W. Then anything from \mathbb{R}^m mapping to b_v will do the trick. And for the uniqueness when A is invertible (or better, when \ker{A}=0) follows immediately. 1 Your claim is correct. Suppose a\in \ell^2, b \in \ell^2 is the vector you conjectured was closest to a (b_i = \max(a_i, 0) for all i), and c is the actual closest nonnegative vector to a in \ell^2. I'll show that c = b. Pick any i for which b_i \ne c_i. Case 1: a_i \ge 0. Then b_i = a_i, and the ith term of the distance d(a,... 0 If you would ask me, it is meant in the sense you mentioned. Since a Hilbert space is isometric isomorphic to its dual space due to the Riesz representation theorem you can identify the functionals in H' with their riesz representers in H. I did that frequently in my bachelor thesis. It is not the most formal way but you can give shorter definitions and ... 1 \langle 3x,3x\rangle=3x doesn't make sense, as the left-hand side is a number and the right-hand side is a function. Perhaps some of the problem is the notation. The correct evaluation is \langle K_3,K_3\rangle =K_3(3) = 9. Note that the RKHS in this case is one dimensional, all of its elements being of the form f(x) = cx =K_c(x) for some c\in\... 0 1) "In particular every bounded operator A: \mathcal{H} \to \mathcal{H} is an unbounded operator". This is confusing at best. Seems to be saying that bounded operators are a subset of unbounded operators, but that is not the usual terminology AFAIK. 2)For example, multiplication by n defined on vectors in l^2(N) with finite numbers of non-zero ... 2 Certainly. Fix \psi\in C^\infty_c with \int\psi=1. Say ||\psi||_2=c. Now say f\in L^2 and \int f=0. Let \epsilon>0. Choose \phi\in C^\infty_c with ||f-\phi||_2<\epsilon. Now$$\left|\int\phi\right| =\left|\int(f-\phi)\right|\le||f-\phi||_2<\epsilon.$$So if \alpha=\int\phi then \int(\phi-\alpha\psi)=0 and$$||f-(\phi-\alpha\... 0 I think your second argument is somewhat going into the wrong direction. You want to show that$H'\subset\mathcal{D}'$, so have to show two things: 1) Given$f\in H'$, you have$f\in\mathcal{D}'$, which follows since$\mathcal{D}$is continuously embedded into$H$(that you still have to show). 2) If two functionals$f,g\in H'$coincide on$\mathcal{D}$, ... 0 You are not allowed to do equip$ \mathcal{D}$with the restriction$ \|\cdot\|_{\mathcal{D}}$of the norm induced by$⟨⋅,⋅⟩_H$, because if you change the norm on$ \mathcal{D}$you change$ \mathcal{D}' $. For example condider$ C^{\infty}_{c}$with the$ L^2 $norm then the dual is$ L^2$. 1 The short answer is:$E^\perp\subset U$implies$U^\perp\supset \overline{E}=E^{\perp\perp}\supset U^\perp$. Let's see what's going on here. First, if$X\subset Y\subset H$, then$X^\perp\supset Y^\perp$just by definition. Applied to the subspaces in question we get$E^{\perp\perp}\supset U^\perp$. Further,$E^{\perp\perp}\$ is a closed subspace containing ...

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To your first question, let $$p=\begin{bmatrix}1&0\\0&0\end{bmatrix},\ \ \ q=\begin{bmatrix}1/2&1/2\\1/2&1/2\end{bmatrix}.$$ Then $$p\vee q=\begin{bmatrix}1&0\\0&1\end{bmatrix},\ \ \ p+q=\begin{bmatrix}3/2&1/2\\1/2&1/2\end{bmatrix},$$ so $$p+q-p\vee q=\begin{bmatrix}1/2&1/2\\1/2&-1/2\end{bmatrix},$$ which is ...

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